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  "body": "\n## Statement\n\nLet $w \u003e 0$, and let $C$ be a [[curve of constant width|B4YD1ZXRVH]] $w$ in the plane.\nThen the [[length|0TH78S7GBZ]] of $C$ is $\\pi w$.\n\n## Remarks\n\nApply the [[Crofton formula|9D5NT00H6G]] to $C$.\nA line that meets the interior of the convex body bounded by $C$ meets $C$ in exactly two points.\nThe lines with normal direction $\\theta$ that meet the body form a band of width $w(\\theta)$.\nSo the length of $C$ is $\\frac{1}{2} \\int_0^\\pi 2 \\, w(\\theta) \\, d\\theta = \\int_0^\\pi w \\, d\\theta = \\pi w$.\nExample: a circle of diameter $w$ has length $\\pi w$, and the Reuleaux triangle of width $w$, made of three arcs of radius $w$ and angle $\\pi/3$, also has length $\\pi w$.\nThe hypothesis of constant width is needed: the square of side $1$ has width $1$ in the direction of its sides, but its perimeter is $4$, not $\\pi$.\n\n## Explanations\n\n### L1 human\n\nAll shapes that have the same width in every direction also have the same perimeter, namely $\\pi$ times the width.\nSo a Reuleaux triangle of width $1$ and a coin of diameter $1$ need exactly the same length of string to go around them.\n\n### L2 human\n\nThe proof shows a more general fact: the perimeter of any plane convex body equals the integral of its width over the directions in $[0, \\pi)$, which is Cauchy's formula.\nBarbier's theorem is the special case of a constant integrand.\nTogether with the isoperimetric inequality it shows that the disc has the largest area among bodies of constant width $w$; the smallest area is the subject of the Blaschke–Lebesgue theorem.\n",
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